The equation for the object's height s at time t seconds after launch is:

s(t) = -4.9t^2 + 19.6t + 58.8, where s is in meters.

What is the starting height?

The starting height is 58.8 m.

What is the maximum height?

We need to find the maximum for s(t) = -4.9t^2 + 19.6t + 58.8.

s max = c - b^2/(4a), where a = -4.9, b = 19.6, c = 58.8

s max = 58.8 - 19.6^2/(4*(-4.9))

s max = 78.4 m

The maximum height is 78.4 m.

How long does it take to reach that height?

t max = -b/(2a), where a = -4.9, b = 19.6

t max = -19.6/(2*(-4.9))

t max = 2 s

It takes 2 seconds to reach that height.

When does the object strike the ground?

The object strikes the ground when s = 0 that is we need to solve the quadratic equation:

-4.9t^2 + 19.6t + 58.8 = 0

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click here to see the equation solved for t

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t = 6 s (We need to ignore the negative solution -2 because the time can only be positive.)

The object will hit the ground after 6 seconds.

What is the height after 3 seconds?

Plug t = 3 s into s(t) = -4.9t^2 + 19.6t + 58.8:

s(3) = -4.9*3^2 + 19.6*3 + 58.8

s(3) = 73.5 m

After 3 seconds the object is 73.5 meters high.

When is the object 45 meters from the ground?

We need to solve the equation s(t) = 45 that is -4.9t^2 + 19.6t + 58.8 = 45 for t:

- 4.9t^2 + 19.6t + 58.8 = 45

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click here to see the equation solved for t

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t = 4.61 s (We need to ignore the negative solution -0.61 because the time can only be positive.)

After 4.61 seconds the object is 45 meters from the ground.